Given points `A(1, 5), B(-3,7) and C(15,9)`
Find the equation of the line through C and parallel to AB.

Answer

To find the equation of the line through point C(15, 9) that is parallel to the line segment AB, we will follow these steps:

### Step 1: Calculate the slope of line AB
The slope (m) of a line through two points (x1, y1) and (x2, y2) is given by the formula:
\[ m = \frac{y2 - y1}{x2 - x1} \]

For points A(1, 5) and B(-3, 7):
- \( x1 = 1, y1 = 5 \)
- \( x2 = -3, y2 = 7 \)

Substituting the values:
\[ m_{AB} = \frac{7 - 5}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2} \]

### Step 2: Use the slope to find the equation of the line through point C
Since the line we want is parallel to AB, it will have the same slope:
\[ m = -\frac{1}{2} \]

The point-slope form of the equation of a line is given by:
\[ y - y1 = m(x - x1) \]

For point C(15, 9):
- \( x1 = 15, y1 = 9 \)

Substituting the values into the point-slope formula:
\[ y - 9 = -\frac{1}{2}(x - 15) \]

### Step 3: Simplify the equation
Distributing the slope on the right side:
\[ y - 9 = -\frac{1}{2}x + \frac{15}{2} \]

Now, add 9 to both sides to isolate y:
\[ y = -\frac{1}{2}x + \frac{15}{2} + 9 \]

Convert 9 to a fraction with a denominator of 2:
\[ 9 = \frac{18}{2} \]

Now combine the fractions:
\[ y = -\frac{1}{2}x + \frac{15}{2} + \frac{18}{2} \]
\[ y = -\frac{1}{2}x + \frac{33}{2} \]

### Step 4: Rearranging to standard form
To write the equation in standard form \( Ax + By + C = 0 \):
Multiply through by 2 to eliminate the fraction:
\[ 2y = -x + 33 \]

Rearranging gives:
\[ x + 2y - 33 = 0 \]

### Final Answer
The equation of the line through point C(15, 9) and parallel to line AB is:
\[ x + 2y - 33 = 0 \]